In a sampling a large number of parts manufactured by a machine , the mean number of defectives in a sample of 20 is 2. Out of 100 such samples ,how many would you expect to contain 3 defective

i)using the Binomial distribution

ii)Poisson distribution

Given, mean number of defectives = 2 = np = 20p.

∴ Probability of a defective part in a single trial is p = 2 / 20 = 0.1 So, q = 1 – p = 0.9 = probability of a non-defective part in a single trial.

∴ Probability of at least three defectives in a sample of 20

= P(X ≥ 3)

= 1– P(X < 3)

= 1 – [P(X = 0) + P(X = 1) + P(X = 2)]

=0.323 .

Therefore, the number of samples having at least three defective parts out of 100 samples

= 100 × 0.323

= 32.3